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LuckyWell [14K]

3 years ago

12

A fair coin is flipped twice. Let X be the number of heads in the two tosses, and Y be the indicator r.v. for the tosses landing

the same way. That is, Y is equal to 1 if the tosses land in the same way, and 0 otherwise. (a) Find the joint pmf of X and Y.

1 answer:

kiruha [24]3 years ago

4 0

Answer:

Step-by-step explanation:

It can be clearly observed that $X\in{0,1,2}$ and $Y\in {0,1}$

Now lets find the probabilities of all possible combinations

$P(X=0, Y=0)=P(Y=0|X=0)P(X=0)=0$

Since it is impossible that the coin ha flipped in different ways given that all the time the tail has fallen

$P(X=1,Y=0)=P(Y=0|X=1)P(X=1)=1\times2\times(\frac{1}{2} )^2$ =1/2

$P(X=2,Y=0)=P(Y=0|X=2)P(X=2)=1\times2\times(\frac{1}{2} )^2$ =1/2

and similarly,

$P(X=0,Y=1)=P(Y=1|X=0)P(X=0)=1\times(\frac{1}{2} )^2= 1/4$

$P(X=1,Y=1)=P(Y=1|X=1)P(X=1)=0$

$P(X=2,Y=1)=P(Y=1|X=2)P(X=2)=1\times(\frac{1}{2} )^2=1/4$

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1) Data given and notation

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$s=4.83$ represent the sample standard deviation

$n=50$ sample size

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